Is ${346641}$ divisible by $9$ ?
A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {346641}= &&{3}\cdot100000+ \\&&{4}\cdot10000+ \\&&{6}\cdot1000+ \\&&{6}\cdot100+ \\&&{4}\cdot10+ \\&&{1}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {346641}= &&{3}(99999+1)+ \\&&{4}(9999+1)+ \\&&{6}(999+1)+ \\&&{6}(99+1)+ \\&&{4}(9+1)+ \\&&{1} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {346641}= &&\gray{3\cdot99999}+ \\&&\gray{4\cdot9999}+ \\&&\gray{6\cdot999}+ \\&&\gray{6\cdot99}+ \\&&\gray{4\cdot9}+ \\&& {3}+{4}+{6}+{6}+{4}+{1} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${346641}$ is divisible by $9$ if ${ 3}+{4}+{6}+{6}+{4}+{1}$ is divisible by $9$ Add the digits of ${346641}$ $ {3}+{4}+{6}+{6}+{4}+{1} = {24} $ If ${24}$ is divisible by $9$ , then ${346641}$ must also be divisible by $9$ ${24}$ is not divisible by $9$, therefore ${346641}$ must not be divisible by $9$.